Problem 271
Question
Find the horizontal and vertical asymptotes. \(f(x)=x-\frac{9}{x}\)
Step-by-Step Solution
Verified Answer
Vertical asymptote at \(x = 0\). No horizontal asymptote.
1Step 1: Identify Horizontal Asymptotes
Horizontal asymptotes occur when \(\lim_{{x \to \infty}} f(x) \) and \(\lim_{{x \to -\infty}} f(x) \)are calculated. For the function \( f(x) = x - \frac{9}{x}, \lim_{{x \to \infty}} \left( x - \frac{9}{x} \right) = \lim_{{x \to \infty}} x - \lim_{{x \to \infty}} \frac{9}{x} = x - 0 = x.\)Since it approaches infinity, there is no horizontal asymptote.
2Step 2: Identify Vertical Asymptotes
Vertical asymptotes occur at values of \( x \) where the function is undefined. The function \( f(x) = x - \frac{9}{x} \) is undefined when the denominator \( x \) is zero. Solve the equation \( x = 0 \) to find the vertical asymptote.Thus, there is a vertical asymptote at \[ x = 0. \]
Key Concepts
Horizontal AsymptotesVertical AsymptotesLimits
Horizontal Asymptotes
Horizontal asymptotes help us understand the behavior of a function as it moves far to the left or right on a graph. They are like road signs showing what direction the function is heading when approaching infinity or negative infinity.
For any function, if you want to find horizontal asymptotes, you calculate
Using the example \( f(x) = x - \frac{9}{x} \), let’s calculate these limits:
Since these limits result in infinity and negative infinity, there is no horizontal asymptote for \( f(x) = x - \frac{9}{x} \).
The rule of thumb is that horizontal asymptotes often look at constants, and our function keeps growing!
For any function, if you want to find horizontal asymptotes, you calculate
\( \lim_{{x \to \infty}} f(x) \): This limit checks what happens as “x” goes very big.\( \lim_{{x \to -\infty}} f(x) \): This limit checks what happens as “x” goes very small.
Using the example \( f(x) = x - \frac{9}{x} \), let’s calculate these limits:
- \( \lim_{{x \to \infty}} \left( x - \frac{9}{x} \right) = \lim_{{x \to \infty}} x - \lim_{{x \to \infty}} \frac{9}{x} = \infty \)
- \( \lim_{{x \to -\infty}} \left( x - \frac{9}{x} \right) = \lim_{{x \to -\infty}} x + \lim_{{x \to -\infty}} \frac{9}{x} = -\infty \)
Since these limits result in infinity and negative infinity, there is no horizontal asymptote for \( f(x) = x - \frac{9}{x} \).
The rule of thumb is that horizontal asymptotes often look at constants, and our function keeps growing!
Vertical Asymptotes
Vertical asymptotes are fascinating because they represent values of "x" that cause a function to become undefined or "blow up," going off to infinity. Think of them as barriers the graph of a function cannot touch.
To discover vertical asymptotes, we pay attention to denominators. Wherever a denominator equals zero and the function isn’t canceled out, that’s a potential vertical asymptote.
For \( f(x) = x - \frac{9}{x} \), \
When sketching a graph, vertical asymptotes are represented by dashed vertical lines. They are warning fences where the function becomes infinitely large.
To discover vertical asymptotes, we pay attention to denominators. Wherever a denominator equals zero and the function isn’t canceled out, that’s a potential vertical asymptote.
For \( f(x) = x - \frac{9}{x} \), \
- Notice the denominator "x" can create a division by zero problem when \( x = 0 \).
- So \( x = 0 \) is a vertical asymptote, meaning the function shoots up to infinity as it gets close to \( x = 0 \).
When sketching a graph, vertical asymptotes are represented by dashed vertical lines. They are warning fences where the function becomes infinitely large.
Limits
Limits are a cornerstone concept in calculus and essential for understanding behaviors at infinity or points of discontinuity. They help determine what value a function approaches as the input gets closer to a certain point or goes towards infinity.
Applying limits in our context helps spot asymptotes and understand function trends.
For our function \( f(x) = x - \frac{9}{x} \), calculating its limits at infinity helped indicate the lack of horizontal asymptotes since:
Similarly, solving for \( x = 0 \) using limits confirmed the vertical asymptote:
By embracing limits, we develop a sense of prediction about where functions stretch or speed off toward. It’s a toolkit for foresight in calculus!
Applying limits in our context helps spot asymptotes and understand function trends.
For our function \( f(x) = x - \frac{9}{x} \), calculating its limits at infinity helped indicate the lack of horizontal asymptotes since:
- \( \lim_{{x \to \infty}} x = \infty \)
- \( \lim_{{x \to -\infty}} x = -\infty \)
Similarly, solving for \( x = 0 \) using limits confirmed the vertical asymptote:
- As \( x \to 0^+ \), \( \frac{9}{x} \to +\infty \).
- As \( x \to 0^- \), \( \frac{9}{x} \to -\infty \).
By embracing limits, we develop a sense of prediction about where functions stretch or speed off toward. It’s a toolkit for foresight in calculus!
Other exercises in this chapter
Problem 270
Evaluate the limit. \(\lim _{x \rightarrow \infty} \frac{2 \sqrt{x}}{x-\sqrt{x}+1}\)
View solution Problem 271
For the following exercises, find the horizontal and vertical asymptotes. $$ f(x)=x-\frac{9}{x} $$
View solution Problem 272
For the following exercises, find the horizontal and vertical asymptotes. $$ f(x)=\frac{1}{1-x^{2}} $$
View solution Problem 272
Find the horizontal and vertical asymptotes. \(f(x)=\frac{1}{1-x^{2}}\)
View solution